Problem: You have found the following ages (in years) of 4 zebras. The zebras are randomly selected from the 48 zebras at your local zoo: $ 18,\enspace 5,\enspace 13,\enspace 8$ Based on your sample, what is the average age of the zebras? What is the variance? You may round your answers to the nearest tenth.
Because we only have data for a small sample of the 48 zebras, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{49} + {36} + {4} + {9}} {{4 - 1}} $ $ {s^2} = \dfrac{{98}}{{3}} = {32.67\text{ years}^2} $ We can estimate that the average zebra at the zoo is 11 years old. There is a variance of 32.67 years $^2$.